Print columns with Awk or Cut?

Question

I'm writing a script that will take a filename as an argument, find a word a specific word at the beginning of each line - the word ATOM, in this case - and print the values from specific columns.

$FILE=*.pdb *

if test $# -lt 1
then
 echo "usage: $0 Enter a .PDB filename"
 exit
fi
if test -r $FILE
then
 grep ^ATOM $FILE | awk '{ print $18 }' | awk '{ print NR $4, "\t" $38,}'
else
 echo "usage: $FILE must be readable"
 exit
fi

I'm having trouble figuring out three problems:

1. How to use awk to print only lines that contain ATOM as the first word
2. How to use awk to print only certain columns from the rows that match the above criteria, specifically columns 2-20 and 38-40
3. How can I indicate this must be a pdb file? *.pdb *

Answer 1

1. That would be

   awk '$1 == "ATOM"' $FILE
   

2. That task is probably better accomplished with cut:

   grep ^ATOM $FILE | cut -c 2-20,38-40
   

3. If you want to ensure that the filename passed as the first argument to your script ends with .pdb: first, please don't (file extensions don't really matter in UNIX), and secondly, if you must, here's one way:

   "${1%%.pdb}" == "$1" && echo "usage:..." && exit 1
   

   This takes the first command-line argument ($1), strips the suffix .pdb if it exists, and then compares it to the original command-line argument. If they match, it didn't have the suffix, so the program prints a usage message and exits with status code 1.

Answer 2

Contrary to the answer, your task can be accomplished with just one awk command. No need grep or cut or ...

if [ $# -lt 1 ];then
 echo "usage: $0 Enter a .PDB filename"
 exit
fi
FILE="$1"
case "$FILE" in
*.pdb )

if test -r $FILE
then 
 # do for 2-20 assuming whites paces as column separators
 awk '$1=="ATOM" && NF>18 { 
   printf "%s ",$2
   for(i=3;i<=19;i++){
     printf "%s ",$i
   }
   printf "%s",$20   
 }' "$FILE"
else
 echo "usage: $FILE must be readable"
 exit
fi
;;
*) exit;;
esac